The equation of the plane which contains the point (2, 0, 1) and perpendicular to the line [tex](x,y, z) = (0, 2, 3) + t\cdot (3, -1, 4)[/tex] is [tex]3\cdot x - y +4\cdot z = 10[/tex].
First, we use the dot product between the vector slope and a vector that pass through and parallel to the plane to derive the equation of the plane:
We determine the vector slope associated with the line:
[tex](x,y, z) = (0, 2, 3) + t\cdot (3, -1, 4)[/tex] (1)
By direct inspection, we see that vector slope is [tex](3, -1, 4)[/tex].
Then, we construct the following vector equation by means of the dot product between the vector slope and a vector passing through and parallel to the plane:
[tex](x-2, y, z-1)\,\bullet\,(3, -1, 4) = 0[/tex]
[tex]3\cdot (x-2) -y +4\cdot (z-1) = 0[/tex]
[tex]3\cdot x -6 -y +4\cdot z -4 = 0[/tex]
[tex]3\cdot x - y +4\cdot z = 10[/tex] (2)
The equation of the plane which contains the point (2, 0, 1) and perpendicular to the line [tex](x,y, z) = (0, 2, 3) + t\cdot (3, -1, 4)[/tex] is [tex]3\cdot x - y +4\cdot z = 10[/tex].
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